Question

find the inverse of the function: $f(x) = 3x^2 + 8$

Explanation:

Step1: Replace \( f(x) \) with \( y \)

We start by writing the function as \( y = 3x^2 + 8 \). This is a standard first step when finding the inverse of a function, as it makes it easier to swap the roles of \( x \) and \( y \) later.

Step2: Swap \( x \) and \( y \)

Now we interchange \( x \) and \( y \) to get \( x = 3y^2 + 8 \). This is based on the definition of an inverse function, where if \( y = f(x) \), then \( x = f^{-1}(y) \), so we swap the variables to solve for the new \( y \) which will be the inverse function.

Step3: Solve for \( y \)

First, subtract 8 from both sides: \( x - 8 = 3y^2 \). Then divide both sides by 3: \( \frac{x - 8}{3}=y^2 \). To solve for \( y \), we take the square root of both sides. Remember that when we take the square root, we get both a positive and a negative solution, so \( y=\pm\sqrt{\frac{x - 8}{3}} \). We can also rationalize the denominator (though it's not necessary, but for a more simplified form) as \( y = \pm\frac{\sqrt{3(x - 8)}}{3}=\pm\frac{\sqrt{3x - 24}}{3} \).

Step4: Replace \( y \) with \( f^{-1}(x) \)

Finally, we replace \( y \) with the inverse function notation \( f^{-1}(x) \), so the inverse function is \( f^{-1}(x)=\pm\sqrt{\frac{x - 8}{3}} \) (or the rationalized form). However, we should note that the original function \( f(x)=3x^2 + 8 \) is not one - to - one over the entire real number line (since it's a quadratic function opening upwards), but if we restrict the domain (for example, to \( x\geq0 \) or \( x\leq0 \)), we would have a one - to - one function and only one of the square root branches would be the inverse. But for the general process of finding the inverse (treating it as a relation), we include both the positive and negative roots.

Answer:

The inverse of the function \( f(x) = 3x^2+8 \) is \( f^{-1}(x)=\pm\sqrt{\frac{x - 8}{3}} \) (or equivalent forms like \( f^{-1}(x)=\pm\frac{\sqrt{3x - 24}}{3} \)). If we consider the function with a restricted domain (e.g., \( x\geq0 \) for the original function, then the inverse would be \( f^{-1}(x)=\sqrt{\frac{x - 8}{3}} \); if \( x\leq0 \), then \( f^{-1}(x)=-\sqrt{\frac{x - 8}{3}} \)).